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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 38101. |
If polaroids are to be used to avoid glares of in coming light then |
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Answer» VISIBILITY will decrease |
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| 38102. |
A loop, made of straight edges has six corners at A(0, 0, 0), B(1, 0, 0), C(1, 1, 0), D(0, 1, 0), E(0, 1, 1) and F(0, 0, 1) a magnetic field B= B_(0) (overset(^)(i) + overset(^)(k)) T is present in the region. Find the flux passing through the loop ABCDEFA? |
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Answer» Solution :Loop ABCDA lie in x-y PLANE WHOSE area vector `A_(1) = L^(2) overset(^)(k)` where ADEFA lie in y-z plane where are vector `A_2 = L^(2) overset(^)(i)` `phi = B.A, A= A_(1) + A_(2) = (L^(2) overset(^)(k) + L^(2) overset(^)(i))` `B = B_(0) ( overset(^) (i) + overset(^)(k) )( L^(2) k + L^(2) overset(^)(i) )= 2 B_(0) L ^2 ` Wb. |
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| 38103. |
The induced emf in a L-R circuit is maximum : |
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Answer» at the TIME of switching on due to high RESISTANCE |
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| 38104. |
In good conductors the gap between the valence band and the conduction band is |
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Answer» zero |
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| 38105. |
In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? |
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Answer» SOLUTION :When slit width .a. is DOUBLED, angular width (and, CONSEQUENTLY, the linear width too) of central diffraction band GIVEN by `theta=pm(lamda)/(a)` is reduced to one-half of its previous value. Consequently, the amplitude of light for central diffraction band will become twice of its previous value and as a result, the new intensity will be 4 times the original value. |
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| 38106. |
After magnetising a substance , by measuring which quantity we can understand the extend to which it is magnetised ? |
| Answer» Solution :The AMOUNT of magnetism attained by a magnetised body is determined by measuring intensity of MAGNETISATION `(BARM)` . This is defined as magnetic MOMENT per unit volume. | |
| 38107. |
(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets ?(b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Figure) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. |
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Answer» Solution :(a) No. We can obtain induced CURRENT by change in magnetic flux linked with coil. Here coil is stationary hence flux will not change. (b) We can not have any induced current in either of TWO given cases because here magnetic flux does not exist. Hence, there can not be any change of magnetic flux. (c) For RECTANGULAR loop, induced emf is `epsilon` = Bvl which can be taken constant till it comes out of the FIELD perpendicularly, because here alongwith B and v, l also remains constant whereas in case of circular loop, length l is initially increasing and later it is decreasing. See the figure given below. Hence, in this case induced emf does not remain constant.  (d) Here, because of motion of two given magnets, magnetic flux towards R.H.S. goes on increasing with time. Hence, according to Lenz.s law, induced current should flow clockwise in the loop containing capacitor, as SEEN from right to left. So electric current should flow anticlockwise i.e. plate A will lose the electrons and plate B will gain the electrons. Such flow of electrons will make plate A positive with respect to plate B. |
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| 38108. |
Statement I: Form factor becomes different for different waveformsof alternating voltage and current. Statement II: The mean value of alternating voltage or current =(2)/(pi)"rms value=1/(sqrt2)"timespeak value" for any wave form. |
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Answer» Statement I is TRUE, statement II is true, statement II is a correct explanation for statement I. |
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| 38109. |
A resistance network is connected to a battery as shown in the figure below. If the intrrnal resistance of the battery is 5omega, then the value of R (in omega) for maximum power delivered to the network is |
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Answer» 2  Now in loop ABCA, USING KVL we have, `- 3Ri_(1)-4R(i_(1)-i_(2))+ 2Ri_(2)= 0` `- 3Ri_(1)-4Ri_(i)+4Ri_(2) + 2Ri_(2)=0` `7i_(1)=6i_(2)` `rArr i_(1)=(6)/(7)i_(2)` Also,`i_(1)+ i_(2)=i` Hence, `((6)/(7)+1)i_(2)=i or i_(2)=(7)/(13)i` `therefore i_(1)=(6)/(7)i_(2)=(6)/(7)XX(7)/(13)i rArr i_(1)= (6)/(13)i` Now from loop ABDA (which includes cell ), we have, `3Ri_(1) +2Ri_(2) = V` (nearly because internal resistance in not taken in ACCOUNT) `3Rxx(6)/(13)i+2Rxx(7)/(13)i=iR_(ed)` `rArr R_(eq)= R ((18 +14)/(13))=(32)/(13)R` For maximum power delivered to circuit, (internal resistancenof source )= external resistance) `rArr 5 = (32R)/(13) or R=(13xx5)/(32)=203 Omega` or `R = 2 Omega` |
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| 38110. |
In 1992, Swiss police arrested two men who were attempting to smuggle osmium out of Eastern Europe for a clan- destine sale. However, by error, the smugglers had picked up ""^(137)Cs. Reportedly, each smuggler was carrying a 1.0 g sample of ""^(137)Cs in a pocket! In (a) bequerels and (b) curies, what was the activity of each sample? The isotope ""^(137)Cs has a half-life of 30.2 y. (The activities of radioisotopes commonly used in hospitals range up to a few millicuries.) |
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| 38111. |
A proton moves horizontally towards a vertical conductor with a uniformly distributed positive charge. It will, undergo |
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Answer» HORIZONTAL DEFLECTION |
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| 38112. |
A capacitor, a resistor and an inductor are joined in series with an ac source. As the frequency of the source is slightly increased from a very low value, the reactance |
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Answer» of the INDUCTOR INCREASES |
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| 38113. |
In a L-C-R circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to ….. |
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Answer» 4 L `therefore L_1C_1=L_2C_2` `therefore LC=L_2(2C)` `therefore L_2=(LC)/(2C)=L/2` |
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| 38114. |
What is the capacitance reactance of a 5mu F capacitor when it is part of a circuit whose frequency is (i) 50 Hz (ii) 10^(6) Hz ? |
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Answer» |
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| 38115. |
A rubber cord of density d, Young's modulus Y and length L is suspended vertically . If the cord extends by a length 0.5 L under its own weight , then L is |
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Answer» `Y/(2dg)` |
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| 38116. |
A ray of light is incident on a glass slab at the polarising angle of58^(@) . Calculate the percentage change in the speed of light in glass. |
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Answer» Solution :Data : ` i_(P) = 58^(@)` ` n = tan i_(P)` ` = tan 58^(@) = 1.6003` ` n = lambda_(a)/lambda_(G)` ` :.lambda_(g)/lambda_(a) = 1/n = 1/(1.6003)` ` :. (lambda_(a)-lambda_(g))/lambda_(a) = (1.6003 - 1)/(1.6003) = (0.6003)/(1.6003)` ` = 0.3753` `{:(log 0.6003,," "bar(1).7784),(log 1.6003,,UL(-0.2041)),(,,ul(" "bar(1).5743)),(ALbar(1).5743,=," "0.3753):}` ` :. ` The percentage change in the velocity of the light in glass is `(lambda_(a)-lambda_(g))/lambda_(a) XX 100 = 37.53 %` |
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| 38117. |
Consider a sinusoidal wave travelling in positive x direction as shown in figure. The wave velocity is 40 cm/s. Find: The velocity of a particle at point P at the instant shown. |
| Answer» SOLUTION :1.26 m/s, DOWNWARD | |
| 38118. |
Consider a sinusoidal wave travelling in positive x direction as shown in figure. The wave velocity is 40 cm/s. Find: The phase difference between points 2.5 cm apart. |
| Answer» SOLUTION :`(5PI)/4 RAD` | |
| 38119. |
Consider a sinusoidal wave travelling in positive x direction as shown in figure. The wave velocity is 40 cm/s. Find : How long it takes for the phase at a given position to change by 60^@ |
| Answer» SOLUTION :`1/60 SEC` | |
| 38120. |
A satellite revolving around the earth in a circular orbit has 4 times the radius of the parking orbit. The time-period of the satellite will be - |
| Answer» Answer :C | |
| 38121. |
In the equation y = 5 sin (0.5t+0.05), representing the wave form of the motion in which y denotes the displacement at the instant t. Which of the following gives the time period of |
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Answer» `2PI` |
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| 38122. |
Substances whose conductivity lies between conductors and insulators are called ? |
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Answer» |
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| 38123. |
A capacitor with initial charge q_0, is discharged through a resistor. What multiple of the time constant tau gives the time the capacitor takes to lose (a) the first 25% of its charge and (b) 50% of its charge? |
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Answer» |
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| 38124. |
Determine the angular rotation velocity of an S_(2) molecule promoted to the first exccited rotational level if the distance between its nuclei is d= 189p m |
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Answer» Solution :In the first EXCITED raotaional level `J=1` so `E_(J)=1xx2( ħ^(2))/(2I)=(1)/(2)I omega^(2)` classically Thus `omega= sqrt(2)( ħ)/(I)` Now `I=Sigma m_(I)r_(i)^(2)=(m)/(2)(d^(2))/(4)+(m)/(2)(d^(2))/(4)=m(d^(2))/(4)` where `m` is the mass of the mole CUB and `r_(i)` is the distance of the atom from the AXIS. Thus `omega=(4sqrt(2) ħ)/(md^(2))= 1.56xx10^(11) rad//s` |
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| 38125. |
A magnet, when suspended in an external magnetic field, has a period of oscillation of 4 s. When it is cut length wise, and suspended in the same magnetic field, the period of vibration will be |
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Answer» `2 SQRT(2) s` |
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| 38126. |
From what the name electricity is coined ? Explain its meaning. |
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Answer» Solution :The name ELECTRICITY is coined from the Greek word ELEKTRON meaning amber. The intrinsic property due to which the OBJECT placed in electric FIELD experiences electric force is called electric charge. It has two types : POSITIVE and negative. |
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| 38127. |
में क्लोरीन परमाणु 71gm के कितने मोल होंगे |
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Answer» 3 |
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| 38128. |
In the following circuit in state of resonance, which statement is correct? |
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Answer» Power FACTOR is `phi` |
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| 38129. |
A particle moves in the X-Y plane under the influence of a force such that its linear momentum isvecp(t) = A[(wedge)icos (kt) -(wedge) sin (kt)] where A and k are constants. The angle between the force and the momentum is: |
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Answer» `0^(@)` `vecf=(dvecp)/(DT)=(d)/(dt)[A(hati coskt-hatj sin kt)]` Thus `vecF.vecp=A k[-veci sin kt-hatj COS kt]` `[A(hati cos kt-hatjsin kt)]=0` The angle between `vecF` and `vecp is 90^(@)` . Hence (d) is the correct choice. |
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| 38130. |
5xx10^(-4) field lines are passing through a coil 1000 turn in certain time interval, the electromotive force of 5V is produced then the time interval will be _____ |
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Answer» 1 s `therefore Deltat=(N DELTAPHI)/epsilon=(1000xx5xx10^(-4))/5`=0.1 s |
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| 38131. |
A coil of 40Omega resistance has 100 turns and radius 6 mm is connected to ammeter of resistance of 160 ohms. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, 32 muCcharge flows through it. The magnetic induction field strength is. |
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Answer» 6.55 T |
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| 38133. |
Pick out the correct statement from the following |
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Answer» Energy RELEASED per unit mass of the reactantis less in case of fusion reaction. |
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| 38134. |
A carbon resistor has 4 bands of white, brown, red and silver colours respectively. The value of resistance is |
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Answer» `(2.2 PM 10%) kOmega` |
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| 38135. |
Solve the above question, if the planes of coil are perpendicular. |
| Answer» Solution :Let a CURRENT i flow in the coil of radius `a_(1)`. The MAGNETIC FIELD at the centre of this coil will now be parallel to the PLANE of smaller coil and HENCE no flux will pass through it, hence M =0 | |
| 38136. |
A given semiconductor has electron concentration of 8xx10^13 "per cm"^3 and a hole concentration of 5xx10^12 per cm^3. What is the resistivity of this sample if the electron mobility is 23,000 cm^2/V and hole mobility is 100 cm^2/V ? |
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Answer» `3.395xx10^(-4)` ohm x CM |
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| 38137. |
In a common-base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 A. The value of the base current amplification factor (beta) will be |
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Answer» 49 `BETA=(I_C)/(I_B)=(5.488)/(0.112)=49` |
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| 38138. |
The ratio of longest wavelength and the shortest wavelength observed in the 5 spectral series of emission spectrum of hydrogen is ....... |
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Answer» `(4)/(3)` `"ln" (1)/(lambda)=R[(1)/(m^(2))-(1)/(n^(2))], m=5 and n=6` `:. (1)/(lambda_(F))=R[(1)/(5^(2))-(1)/(6^(2))]` `:. (1)/(lambda_(F))=(11R)/(900)` `:. (1)/(lambda_(F))=(900)/(11R)...(1)` For shortest wavelength in LYMAN series m=1 and `n=oo` `:. (1)/(lambda_(L))=R[(1)/(1^(2))-(1)/(oo^(2))]=(R)/(1)` ` :. (1)/(lambda_(L))=(1)/(R) ...(2)` `RARR` Ratio `=(lambda_(F))/(lamda_(L))=(900)/(11R)xx(R)/(1)=(900)/(11)` |
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| 38139. |
Half-life of a radioactive sample is 5 days. What time is taken for 7/8th part of the sample of decay ? |
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Answer» 3.4 days |
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| 38140. |
Statement -1: A nucleus having energy E_1 decays by beta^(-) emission to daughter nucleus having energy E_1 but the beta^(-) rays are emitted with a continuous energy spectrum having end point energy E_(1)-E_(2). Statement - 2: To conserve energy and momentum in beta -decay at least three particles must take part in the transformation. |
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Answer» Statement-1 is correct but Statement-2 is not correct. Statement-2 : For energy conservation and momentum at least three particles daughter NUCLEUS, `+beta^(-1)` and antineutron. |
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| 38141. |
In the circuit given below, both batteries are ideal. Emf E_(1) of battery 1 has a fixed value,but emf E_(1) of battery 2 can be varied between 1.0 V and 10.0 and 10.0 V. The graph gives the currents through the two batteries as a function of E_(2), but are not marked as which plot corresponds to which battery. But for both ploth, current is assumed to be negative when the direction of the current through the cattery is oppositi the direction of that battery's emf. (Direction of emf is from negative to positive) The resistance R_(1) has value |
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Answer» `10 Omega` |
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| 38142. |
A coil of inductance 0.50 H and resistance 100omegais connected to a 240 V, 50 Hz a.c. supply What is the time lag between voltage maximum and current maximum . |
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Answer» Solution :PHASE DIFFERENCE `PHI` is given by `tan phi = (omegaL)/R = (2pi xx 50 xx 0.5) /100 = 1.571 [phi = tan ^(-1) (1.571) = 57.5^@ V]` |
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| 38143. |
A coil of inductance 0.50 H and resistance 100omegais connected to a 240 V, 50 Hz a.c. supply What is maximum current in the coil ? |
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Answer» SOLUTION :Impedance Z = `sqrt(R^2 + omega^2 L^2)` `=sqrt((1000)^2 - 1 (2PI XX 50 xx 0.5)^2 omega)`= 186.21 `omega` |
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| 38144. |
The angular frequency of a fan increases from 30 rpm to 60 rpm in pis. A dust particle is present at a distance of 20 cm from axis of rotation. The tangential acceleration of the particle in pis is |
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Answer» `0.8 MS^(-2)` |
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| 38145. |
A cell of emf 'E' and internal resistance r connected in the secondary gets balanced against length l of potentiometer wire. If a resistance 'R' is connected in parallel with the cell, then the new balancing length for the cell will be |
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Answer» `((R)/(R-r))L` |
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| 38146. |
In electric dipole is placed in an electric field of a point charge, then........ |
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Answer» (A) the resultant force ACTING on the dipole is always zero. `therefore`Electric field at the position of +q and - q are not equal. Hence, resultant force is never zero, so option (A) and (B) are wrong. If dipole becomes parallel to electric field `vecE`,then angle between `vecp` and `vecE` will be 0°, so the torque acting on dipole `tau = pE sin 0^(@) =0`, but if `phi ne 0^(@)`,the torque will not be zero, so option (C) is correct. |
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| 38147. |
The S.I. units of centrifugal force is |
| Answer» Answer :D | |
| 38148. |
The central band in interference pattern is : |
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Answer» ALWAYS dark |
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| 38149. |
On a railway track of radius of curvature 1600 m. If the distance between two trackes is 1.8 m, then the elevation of the outer track above the inner track will be(g=10m//s^(2)) |
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Answer» `0.450 m` Where l is DISTANCE between TWO tracks. |
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| 38150. |
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite sign and of magnitude 17.0 xx 10^(-11)C//m^2. What is E (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? |
Answer» SOLUTION :`sigma=1.70 xx 10^(-11) CM^(-2)`  `a. E_(1)=(sigma)/(2epsi_(0)) -(sigma)/(2epsi_(0))=0` b. `E_(2)=(sigma)/(2epsi_(0)) -(sigma)/(2 epsi_(0))=0` `c. E_(3)=sigma/epsi_(0) =(1.7 xx 10^(-11))/(8.85 xx 10^(-12)) =1.9 NC^(-1)` |
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